Force on a bead due to a rotating wire

[Pushoam]

Homework Statement

A bead slides without friction on a horizontal rigid wire rotating at
constant angular speed ω. The problem is to find the force exerted on
the bead by the wire. Neglect gravity.

Homework Equations

The Attempt at a Solution

The physical force acting on the bead is the force exerted by the rigid wire.Let's call this force $\vec F_{ph}$.
Wrt. inertial frame,
$\vec F_{ph} = m { a_r } {\hat r } + m { a_{\theta}} {\hat {\theta} }$

constant angular speed ω gives $\ddot {\theta} = 0$
$a_r = \ddot r - r {\dot {\theta}}^2$
$a_{\theta} = 2{ \dot r }{\dot {\theta}} + r {\ddot {\theta}} = 2{ \dot r }{\dot {\theta}}$
So,
$\vec F_{ph} =$ m( ($\ddot r - r { {\dot {\theta}}^2 }$) ${\hat r} + 2 { \dot r } {\dot {\theta}} {\hat {\theta}}$ )

Is this correct?

Please, tell me where is the mistake in Latex code? I edited and corrected the code.

 

[TSny]

The physical force acting on the bead is the force exerted by the rigid wire. Let's call this force $\vec F_{ph}$ .
Wrt. inertial frame,
$\vec F_{ph} = ma_r \hat r + ma_{\theta} {\hat {\theta}}$

constant angular speed ω gives $\ddot {\theta} = 0$
$a_r = \ddot r - r {\dot {\theta}}^2$
$a_{\theta} = 2{ \dot r }{\dot {\theta}} + r {\ddot {\theta}} = 2{ \dot r }{\dot {\theta}}$
So,
$\vec F_{ph}$ = m ( $(\ddot r - r {\dot {\theta}}^2 ) \hat r + 2{ \dot r }{\dot {\theta}} {\hat {\theta}}$ )

Is this correct?

It's correct so far, but there's quite a bit more you can do that will lead to an explicit expression for the force as a function of time.

 

[Pushoam]

I corrected the Latex code.

 

[Pushoam]

there's quite a bit more you can do

I have no more information about $\dot r$ or $\ddot r$. So, what more could be done?

 

[TSny]

I have no more information about $\dot r$ or $\ddot r$. So, what more could be done?

The wire is frictionless.

 

[Pushoam]

The wire is frictionless.

This means that $\vec F_{ph}$ has no component in the $\hat r$ direction.

[ Earlier, I thought that there must be some force to provide $\ddot r$. So, I was stuck. I often do this mistake. Thanks for pointing out that there is no friction. ]

So, $a_r = \ddot r - r {\dot {\theta}}^2 = 0$
$\int_{t_i}^{t_f} \left ({\ddot r} - r {\dot {\theta}}^2 \right ) \, dt = 0$
${\dot r}_f - {\dot r }_i = r \left (t_f - t_i \right ) {\dot {\theta}}^2$

Taking ${\dot r }_i = t_i =0$

${\dot r}_f = r t_f {\dot {\theta}}^2$

${\dot r} = r t {\dot {\theta}}^2$

So,
$\vec F_{ph} = 2 m { \dot r } {\dot {\theta}} {\hat {\theta}} = 2 m r t {\dot {\theta}}^3 {\hat {\theta}}$

Is this correct?

I think the following step is wrong as r itself is a function ot t.
${\dot r}_f - {\dot r }_i = r \left (t_f - t_i \right ) {\dot {\theta}}^2$

 

[TSny]

This means that $\vec F_{ph}$ has no component in the $\hat r$ direction.
So, $a_r = \ddot r - r {\dot {\theta}}^2 = 0$

Yes. Can you solve this differential equation?

$\int_{t_i}^{t_f} \left ({\ddot r} - r {\dot {\theta}}^2 \right ) \, dt = 0$
${\dot r}_f - {\dot r }_i = r \left (t_f - t_i \right ) {\dot {\theta}}^2$

You treated $r$ as constant when integrating $r {\dot {\theta}}^2$. But $r$ is some function of time that you are trying to find.

 

[Pushoam]

$a_r = \ddot r - r {\dot {\theta}}^2 = 0$
$\frac{d{\dot r}} {dt} = \frac {d{\dot r}}{dr} \frac {dr} {dt} = r {\dot {\theta}}^2$
$\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}$
Integrating both sides with limit $\dot r$ → 0 to $\dot r$ and r →0 to r
$\frac {{\dot r}^2} 2 = \frac {r^2 {{\dot {\theta}}^2} }{2}$
$\dot r = r \dot {\theta}$

Is this correct so far?

$\vec F_{ph} = 2 m { \dot r } {\dot {\theta}} {\hat {\theta}} = 2 m r {\dot {\theta}}^2 {\hat {\theta}}$

 

[Pushoam]

It's correct so far, but there's quite a bit more you can do that will lead to an explicit expression for the force as a function of time.

The question doesn't say to express the force as a function of time?
So, how do you get to know beforehand that the force should be expressed as a function of time?
Is it because the only one independent variable in this problem is time?

 

[haruspex]

The question doesn't say to express the force as a function of time?
So, how do you get to know beforehand that the force should be expressed as a function of time?
Is it because the only one independent variable in this problem is time?

We cannot tell if you posted the whole problem as given to you.
Clearly the force will depend on r and r will depend on t. Is there any hint as to whether you are to find the force as a function of r or of t (or...)?

 

[Pushoam]

We cannot tell if you posted the whole problem as given to you.
Clearly the force will depend on r and r will depend on t. Is there any hint as to whether you are to find the force as a function of r or of t (or...)?

Actually , this is an example given in the book. But, I wonder how can one know that the force will depend on r before calculating it?

To solve the same question w.r.t. wire frame i.e. non- inertial frame,

we have,
$\vec F_{n-in }= \vec F_{ph} + \vec F_{pseudo}$

where $\vec F_{n-in }$ is net force acting on the system measured wrt. the non - inertial frame
$\vec F_{ph }$ is the net physical force acting on the system measured wrt. the inertial frame
$\vec F_{pseudo }$ is pseudo force acting on the system

Here,
$\vec F_{ph }$ is the contact force exerted by the wire on the bead
$\vec F_{n-in } = m \vec a_{n-in }$
$\vec a_{n-in } = \ddot r {\hat r}$
In this frame, the bead is moving along one direction. Now, How to specify this direction? Should I say that $\hat r$ is constant here?


$\vec F_{pseudo } = -m \vec \omega \times \vec \omega \times \vec r - 2 m \vec \omega \times \vec v_{n-in}$

where $\vec \omega$ is the angular velocity with which the non -inertial frame (here wire) is rotating
$\vec v_{n-in}$ is the velocity of the system wrt. non-inertial frame
Here, $\vec v_{n-in} = \dot r \hat r$

So, now using $\vec F_{n-in }= \vec F_{ph} + \vec F_{pseudo}$ we have,

$m \ddot r~ {\hat r} = \vec F_{ph} - m \vec \omega \times \vec \omega \times \vec r - 2 m \vec \omega \times\dot r \hat r$
$m \ddot r~ {\hat r} = \vec F_{ph} + m { \omega }^2 r~ \hat r - 2 m \omega \dot r~ \hat \theta$

From the above equation, it's clear that $\vec F_{ph}$ can have components only in $\hat r$ or $\hat \theta$ direction.
$\vec F_{ph}$ is a contact force. Its component in $\hat r$ direction is known as friction. And it's given that friction is 0. Thus, the direction of $\vec F_{ph}$ is known from the equation of motion.
So, we have $\vec F_{ph} = 2 m \omega \dot r \hat \theta$

Considering eqn. of motion in $\hat r$ direction, we have the following differential equation,
$\ddot r = { \omega }^2 r$

now , follow post # 8.

 

[haruspex]

$\dot r = r \dot {\theta}$

Is this correct so far?

Looks ok.
Now solve that to get r as a function of theta (and hence of t).

 

[Pushoam]

Looks ok.
Now solve that to get r as a function of theta (and hence of t).

$\dot r = r\dot \theta$
$\frac {dr} {dt} =r \frac {d \theta} {dt}$
Integrating both sides wrt t,
$\int \frac 1 r \frac {dr} {dt} \, dt =\int \frac {d \theta} {dt} \, dt$
$\ln r = \theta + C$
$r =A~ e^{\theta} = A~ e^{\omega t}$

$\vec F_{ph} = 2 m { \dot r } {\dot {\theta}} {\hat {\theta}} = 2 m \omega ^2 ~ A e^{\omega t} ~{\hat {\theta}}$

 

[Pushoam]

In this frame, the bead is moving along one direction. Now, How to specify this direction? Should I say that $\hat r$ is constant here?

What about this?

 

[TSny]

$a_r = \ddot r - r {\dot {\theta}}^2 = 0$
$\frac{d{\dot r}} {dt} = \frac {d{\dot r}}{dr} \frac {dr} {dt} = r {\dot {\theta}}^2$
$\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}$

OK.

Integrating both sides with limit $\dot r$ → 0 to $\dot r$ and r →0 to r
$\frac {{\dot r}^2} 2 = \frac {r^2 {{\dot {\theta}}^2} }{2}$
$\dot r = r \dot {\theta}$

This is not generally true. When you integrated, your lower limits imply that $\dot r = 0$ at $r = 0$. But this was not given in the problem as either an initial condition or a boundary condition. (If the particle starts at $r = 0$ with $\dot r = 0$ then it would remain at rest there.)

Whether you want the force as a function of $t$ or as a function of $r$, the answer will depend on initial conditions (or boundary conditions).

 

[Pushoam]

This is not generally true. When you integrated, your lower limits imply that $\dot r = 0$at r=0. But this was not given in the problem as either an initial condition or a boundary condition. (If the particle starts at r = 0 with $\dot r = 0$ then it would remain at rest there.)

Yes, what I did is wrong.
$a_r = \ddot r - r {\dot {\theta}}^2 = 0$
The initial condition which I took makes a_r 0.
Hence, If the particle starts at r = 0 with $\dot r = 0$ then it would remain at rest there wrt. the wire frame.

$a_r = \ddot r - r {\dot {\theta}}^2 = 0$
$\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}$
Integrating both sides with limit $\dot r$→ v_i to $\dot r$and r →r_i to r,
$\frac {\{{\dot r}^2 - v_i^2\}} 2 = \frac {\{r^2 - r_i^2 \} {{\dot {\theta}}^2} }{2}$
$\dot r = \sqrt{ \frac {v_i^2} 2 + \frac {\{r^2 - r_i^2 \} {{\dot {\theta}}^2} }{2} }$

Is this correct so far?

But, then, I don't get r(t) as an exponential function of t.

 

[TSny]

$a_r = \ddot r - r {\dot {\theta}}^2 = 0$
$\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}$
Integrating both sides with limit $\dot r$→ v_i to $\dot r$and r →r_i to r,
$\frac {\{{\dot r}^2 - v_i^2\}} 2 = \frac {\{r^2 - r_i^2 \} {{\dot {\theta}}^2} }{2}$

Yes. Note the 2's will cancel.

But, then, I don't get r(t) as an exponential function of t.

If you want $r$ as a function of time, go back to the differential equation

$\ddot r - r {\dot {\theta}}^2 = 0$

or

$\ddot r - \omega^2 r = 0$

This is a second order differential equation for $r(t)$. If you have studied differential equations,you should find this equation to be easy to solve

 

[Pushoam]

Note the 2's will cancel.

So,$\dot r = \sqrt{ {v_i^2} + {\{r^2 - r_i^2 \} {{\dot {\theta}}^2} }}$
$\frac {dr} {dt} =\dot \theta \sqrt{c^2 + r^2}$, where $c^2 = \frac {{v_i}^2} {{\dot \theta}^2} -{ r_i}^2$
Integrating both sides wrt t gives,
$\int_{r_i} ^{r_f}\frac 1 {\sqrt{c^2 + r^2} } \, dr = \int_{t_i}^{t_f} \dot \theta\, dt$
$\frac {r_f} {\sqrt {{r_f}^2 + c^2} } - \frac {r_i} {\sqrt {{r_i}^2 + c^2} } =\dot \theta \{ t_f - t_i\}$

Is this correct so far?
Taking c² =0 and r_i=0, I should get the old result. But I don't get so. So, is there anything wrong here?

 

[TSny]

If you want $r(t)$, it will be much easier to solve $\ddot r - \omega^2 r= 0$ directly rather than using $\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}$.

 

[Pushoam]

If you want $r(t)$, it will be much easier to solve $\ddot r - \omega^2 r= 0$ directly rather than using $\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}$.

Yes, that I know. But, I want to know : is there anything wrong with the way I am solving it here?

 

[TSny]

Yes, that I know. But, I want to know : is there anything wrong with the way I am solving it here?

No, it should work. But you haven't integrated correctly in post #18.

 

[Pushoam]

$\int_{r_i} ^{r_f}\frac 1 {\sqrt{c^2 + r^2} } \, dr = \int_{t_i}^{t_f} \dot \theta\, dt$
$\int_{r_i} ^{r_f}\frac 1 {\sqrt{c^2 + r^2} } \, dr = I$
Taking $r = c~ {\tan\alpha}$
$\frac {dr}{d \alpha} =c ~ {\sec ^2 \alpha}$
$dr ={ {\sec ^2 \alpha}}~ d \alpha$
$\int_{r_i} ^{r_f} { \frac 1 {\sqrt{c^2 + r^2} } } \, dr = \int_{\alpha_i} ^{\alpha_f} { \frac {c~ {\sec^2 \alpha} } {c ~\sec \alpha }} \, d \alpha$
$= \int_{\alpha_i} ^{\alpha_f} \sec \alpha \, d \alpha = \ln \{\sec \alpha + \tan \alpha\}|_{\alpha_i} ^{\alpha_f}$
$= \ln { \frac {\frac { {\sqrt{c^2 + r_f^2} } + r_f} c } {\frac { {\sqrt{c^2 + r_i^2} } + r_i} c } }$

$\ln {\frac { {\sqrt{c^2 + r_f^2} } + r_f} { {\sqrt{c^2 + r_i^2} } + r_i}} = \dot \theta\{ t_f - t_i\}$

${ {\sqrt{c^2 + r_f^2} } + r_f} = D e^{ \dot \theta { t_f}}$ , where D is an appropriate constant
${\sqrt{c^2 + r_f^2} } = D e^{ \dot \theta { t_f } } - r_f$
On squaring both sides,
$r_f ^2 + c^2 = G e^{ 2 \dot \theta { t_f } } + r_f^2 - 2 r_f D e^{ \dot \theta{ t_f } }$
$r_f = H e^{ - \dot \theta { t_f } } + G e^{ \dot \theta{ t_f } }$ where c,G,H are appropriate constants.

 

[TSny]

OK. It looks good to me.

 

[Pushoam]

Thanks.
It's done.